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Vsevolod [243]
3 years ago
14

What can the student predict if she finds the mass of the whole candy bar, including the wrapper?

Chemistry
1 answer:
Vesna [10]3 years ago
6 0

Answer:

A student can predict Amount of calories present if she finds the mass of whole candy bar,including the wrapper.

<u>Explanation:</u>

Based on the mass of a whole candy bar a student can predict the calories contained in the candy bar. If the mass of candy bar is more than it means it has more calories and if the candy bar has less mass than it shows the presence of fewer calories.

More calories mean more sugar content in the candy bar. With sugar, the candy bar has carbohydrates also. The sugar present in the bar gives us energy.

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A sample of an Iron Oxalato complex salt weighting 0.145 grams requires 34.88 mL of 0.015 M KMnO4 to turn the solution a very li
Amanda [17]

Answer:

5.2 ×10^-4 moles

Explanation:

Equation of reaction:

5C2O4^2- + 2MnO4^- +. 6H^+. ----------> 10CO2. +. 8H2O. + 2Mn^2+

From the information provided in the question:

Volume of potassium permanganate= 34.88ml

Concentration of potassium permanganate= 0.015M

Amount potassium permanganate= 0.015 × 34.88/1000= 5.2 ×10^-4 moles

6 0
3 years ago
Read 2 more answers
If 25.3 grams of mercury(II) oxide react to form 23.4 grams of mercury, how many grams of oxygen must simultaneously be formed?
sergij07 [2.7K]

Answer : The mass of oxygen formed must be 3.8 grams.

Explanation :

Law of conservation of mass : It states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The balanced chemical reaction will be,

2HgO\rightarrow 2Hg+O_2

According to the law of conservation of mass,

Total mass of reactant side = Total mass of product side

Total mass of 2HgO = Total mass of 2Hg+O_2

As we are given :

The mass of HgO = 25.3 grams

The mass of Hg = 23.4 grams

So,

2\times 25.3g=2\times 23.4g+\text{Mass of }O_2

50.6g=46.8g+\text{Mass of }O_2

\text{Mass of }O_2=3.8g

Therefore, the mass of oxygen formed must be 3.8 grams.

6 0
3 years ago
Which animals use the specialized roles of diver and barrier while participating in cooperative hunting?
amm1812
I believe the answer is The Dolphins.
In their Cooperative hunting, The dolphins requires five of its members to form of barrier while one or two of them are herding (Driving) their preys toward the barriers. And each dolphins will always play the same role in their group just like what we see in a football team.
3 0
3 years ago
Read 2 more answers
Cual es la nomenclatura stock/tradicional/sistematica del oxigeno????
ANEK [815]

Answer:

english pls so i can answer

7 0
3 years ago
A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of
Bad White [126]

Answer:

a) T_{2} = 360.955\,K, P_{2} = 138569.171\,Pa\,(1.386\,bar), b) T_{2} =  347.348\,K, V_{2} = 0.14\,m^{3}

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}

The number of moles of the ideal gas is:

n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}

n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}

n = 5.541\,mol

The final temperature is:

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}

T_{2} = 360.955\,K

The final pressure is:

P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}

P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)

P_{2} = 138569.171\,Pa\,(1.386\,bar)

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}

T_{2} =  347.348\,K

The final volume is:

V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}

V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})

V_{2} = 0.14\,m^{3}

4 0
4 years ago
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